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[MUSIC PLAYING]
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PROFESSOR: Last time, we began
a discussion of feedback.
11
00:00:59,530 --> 00:01:02,240
And I briefly introduced a
number of applications.
12
00:01:02,240 --> 00:01:05,970
For example, the use of feedback
in compensating for
13
00:01:05,970 --> 00:01:08,400
non-ideal elements
in a system.
14
00:01:08,400 --> 00:01:12,900
And also, as another example,
the use of feedback in
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00:01:12,900 --> 00:01:16,050
stabilizing unstable systems.
16
00:01:16,050 --> 00:01:19,680
In this final lecture, what
I'd like to do is focus in
17
00:01:19,680 --> 00:01:22,980
more detail on the use
of feedback to
18
00:01:22,980 --> 00:01:25,930
stabilize unstable systems.
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00:01:25,930 --> 00:01:29,610
And the specific context in
which I'd like to do that is
20
00:01:29,610 --> 00:01:32,640
in the context of the
inverted pendulum.
21
00:01:32,640 --> 00:01:36,230
Now, I've referred to the
inverted pendulum in several
22
00:01:36,230 --> 00:01:37,610
past lectures.
23
00:01:37,610 --> 00:01:42,130
And, basically, as everyone
realizes, a pendulum is
24
00:01:42,130 --> 00:01:45,460
essentially a rod with a
weight on the bottom.
25
00:01:45,460 --> 00:01:48,130
And, naturally, an inverted
pendulum is just
26
00:01:48,130 --> 00:01:49,700
that, upside down.
27
00:01:49,700 --> 00:01:53,740
And so it's essentially a
rod that's top-heavy.
28
00:01:53,740 --> 00:01:58,840
And the idea with the inverted
pendulum is, although by
29
00:01:58,840 --> 00:02:03,270
itself it's unstable around a
pivot point at the bottom, the
30
00:02:03,270 --> 00:02:07,840
idea is to apply an external
input, an acceleration,
31
00:02:07,840 --> 00:02:09,509
essentially to keep
it balanced.
32
00:02:09,509 --> 00:02:14,300
And we've all probably done
this type of thing at some
33
00:02:14,300 --> 00:02:17,970
point in our lives, either
childhood or not.
34
00:02:17,970 --> 00:02:22,930
And so that's the system that
I'd like to analyze and then
35
00:02:22,930 --> 00:02:25,240
illustrate in this lecture.
36
00:02:25,240 --> 00:02:29,630
Now, the context in which we'll
do that is not with my
37
00:02:29,630 --> 00:02:34,360
son's horse, but actually with
an inverted pendulum which is
38
00:02:34,360 --> 00:02:35,830
mounted on a cart.
39
00:02:35,830 --> 00:02:39,220
And first I'd like to go through
an analysis of it, and
40
00:02:39,220 --> 00:02:43,110
then we'll actually see
how the system works.
41
00:02:43,110 --> 00:02:47,950
So basically, then, the system
that we're talking about is
42
00:02:47,950 --> 00:02:51,940
going to be a system which
is a movable cart.
43
00:02:51,940 --> 00:02:56,140
Mounted on the cart is a rod
with a weight on the top.
44
00:02:56,140 --> 00:02:58,910
And this then becomes the
inverted pendulum.
45
00:02:58,910 --> 00:03:05,140
And the cart has an external
acceleration applied to it,
46
00:03:05,140 --> 00:03:09,370
which is the input that we
can apply to the system.
47
00:03:09,370 --> 00:03:13,220
And then, in general, we can
expect some disturbances.
48
00:03:13,220 --> 00:03:17,670
And I'm going to represent the
disturbances in terms of an
49
00:03:17,670 --> 00:03:21,860
angular acceleration, which
shows up around the weight at
50
00:03:21,860 --> 00:03:24,550
the top of the rod.
51
00:03:24,550 --> 00:03:28,070
So if we look at the system,
then, as I've kind of
52
00:03:28,070 --> 00:03:33,840
indicated it here, basically
we have two inputs.
53
00:03:33,840 --> 00:03:37,580
We have an input, which are
the external disturbances,
54
00:03:37,580 --> 00:03:41,580
which we can assume that we
have no control over.
55
00:03:41,580 --> 00:03:45,760
And then there is the
acceleration that's apply to
56
00:03:45,760 --> 00:03:50,620
the cart externally, or in the
case of balancing my son's
57
00:03:50,620 --> 00:03:53,480
horse, it's the movement
of my hand.
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00:03:53,480 --> 00:03:57,760
And then, through the system
dynamics, that ends up
59
00:03:57,760 --> 00:04:02,680
influencing the angle
of the rod.
60
00:04:02,680 --> 00:04:06,680
And we'll think of the
angle of the rod
61
00:04:06,680 --> 00:04:08,760
as the system output.
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00:04:08,760 --> 00:04:14,250
And basically the idea, then,
is to try to keep that angle
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00:04:14,250 --> 00:04:19,680
at 0 by applying the appropriate
acceleration.
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00:04:19,680 --> 00:04:24,360
Now, as I've mentioned several
times previously, if we know
65
00:04:24,360 --> 00:04:29,000
exactly what the system dynamics
are, and if we know
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00:04:29,000 --> 00:04:31,910
exactly what the external
disturbances are, then
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00:04:31,910 --> 00:04:36,030
theoretically we can choose an
acceleration for the cart
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00:04:36,030 --> 00:04:40,430
which will exactly balance the
system, or balance the rod.
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00:04:40,430 --> 00:04:44,390
However, since the system is
inherently unstable, any
70
00:04:44,390 --> 00:04:48,620
deviation from that model-- in
particular, any unanticipated
71
00:04:48,620 --> 00:04:50,580
external disturbances--
72
00:04:50,580 --> 00:04:52,250
will excite the instability.
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00:04:52,250 --> 00:04:55,460
And what that means is that
the rod will fall.
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00:04:55,460 --> 00:05:01,710
So the idea, then, is, as we'll
see, to stabilize the
75
00:05:01,710 --> 00:05:04,950
system by placing feedback
around it.
76
00:05:04,950 --> 00:05:09,190
In particular through a
measurement of the output
77
00:05:09,190 --> 00:05:14,500
angle, processed through some
appropriate feedback dynamics,
78
00:05:14,500 --> 00:05:18,580
using that to control the
acceleration of the cart.
79
00:05:18,580 --> 00:05:22,160
And if we choose the feedback
dynamics correctly, then, in
80
00:05:22,160 --> 00:05:27,030
fact, we can end up with a
stable system, even though the
81
00:05:27,030 --> 00:05:29,760
open-loop system is unstable.
82
00:05:29,760 --> 00:05:34,450
Well, as a first step, let's
analyze the system in its
83
00:05:34,450 --> 00:05:35,880
open-loop form.
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00:05:35,880 --> 00:05:40,090
In particular, let's analyze the
system without feedback,
85
00:05:40,090 --> 00:05:44,390
demonstrate that, indeed, it is
unstable, and then see how
86
00:05:44,390 --> 00:05:47,640
we can stabilize it
using feedback.
87
00:05:47,640 --> 00:05:50,770
So the system, then, is--
88
00:05:50,770 --> 00:05:55,610
as we've indicated, the
variables involved are the
89
00:05:55,610 --> 00:06:00,460
measured angle, which represents
the output, the
90
00:06:00,460 --> 00:06:04,680
angular acceleration due to
external disturbances, and
91
00:06:04,680 --> 00:06:07,260
then there is the
applied external
92
00:06:07,260 --> 00:06:09,720
acceleration on the cart.
93
00:06:09,720 --> 00:06:13,530
So these represent the
variables, with L representing
94
00:06:13,530 --> 00:06:17,300
the length of the rod, and s of
t I indicate here, which we
95
00:06:17,300 --> 00:06:20,430
won't really be paying attention
to, as the position
96
00:06:20,430 --> 00:06:22,230
of the cart.
97
00:06:22,230 --> 00:06:28,040
And without going into the
details specifically, what
98
00:06:28,040 --> 00:06:30,830
we'll do is set up the equation,
or write the
99
00:06:30,830 --> 00:06:34,290
equation, in terms of balancing
the acceleration.
100
00:06:34,290 --> 00:06:38,080
And if you go through that
process, then the basic
101
00:06:38,080 --> 00:06:41,890
equation that you end up with
is the equation that I
102
00:06:41,890 --> 00:06:43,730
indicate here.
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00:06:43,730 --> 00:06:48,710
And this equation, then, tells
us how the basic forces, or
104
00:06:48,710 --> 00:06:52,330
accelerations, are balanced,
where this is the second
105
00:06:52,330 --> 00:06:55,900
derivative of the angle.
106
00:06:55,900 --> 00:06:59,590
And then we have the
acceleration due to gravity,
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00:06:59,590 --> 00:07:05,080
reflected through the angular
acceleration times the sine of
108
00:07:05,080 --> 00:07:09,190
the angle, the angular
acceleration due to the
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00:07:09,190 --> 00:07:12,960
external disturbances, and
finally, the angular
110
00:07:12,960 --> 00:07:16,060
acceleration due to the
motion of the cart.
111
00:07:16,060 --> 00:07:19,730
Now, this equation, as it's
written here, is a nonlinear
112
00:07:19,730 --> 00:07:23,830
equation in the angle
theta of t.
113
00:07:23,830 --> 00:07:28,100
And what we'd like to do is
linearize the equation.
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00:07:28,100 --> 00:07:31,160
And we'll linearize the
equation by making the
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00:07:31,160 --> 00:07:35,670
assumption that the angle is
very small, close to 0.
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00:07:35,670 --> 00:07:38,200
In other words, what we'll
assume is that we're able to
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00:07:38,200 --> 00:07:40,970
keep the rod relatively
vertical.
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00:07:40,970 --> 00:07:44,180
And our analysis, because
we're linearizing the
119
00:07:44,180 --> 00:07:46,850
equations, will obviously
depend on that.
120
00:07:46,850 --> 00:07:51,770
So making the assumption that
the angle is small, we then
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00:07:51,770 --> 00:07:56,070
assume that the sine of the
angle is approximately equal
122
00:07:56,070 --> 00:08:01,050
to the angle, and the cosine of
the angle is approximately
123
00:08:01,050 --> 00:08:02,480
equal to 1.
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00:08:02,480 --> 00:08:04,980
That will then linearize
this equation.
125
00:08:04,980 --> 00:08:10,230
And the resulting equation,
then, in its linearized form,
126
00:08:10,230 --> 00:08:13,380
is the one that I
indicate here.
127
00:08:13,380 --> 00:08:20,000
So we have an equation of
motion, which linearizes the
128
00:08:20,000 --> 00:08:22,260
balance of the accelerations.
129
00:08:22,260 --> 00:08:24,730
And what we see on the
right-hand side of the
130
00:08:24,730 --> 00:08:29,970
equation is the combined inputs
due to the angular
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00:08:29,970 --> 00:08:33,659
acceleration of the rod and the
acceleration of the cart.
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00:08:33,659 --> 00:08:37,950
And on the left-hand side of
the equation, we have the
133
00:08:37,950 --> 00:08:42,419
other forces due to the
change in the angle.
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00:08:42,419 --> 00:08:44,990
So this is, then, the
differential equation
135
00:08:44,990 --> 00:08:49,700
associated with the open-loop
system, the basic dynamics of
136
00:08:49,700 --> 00:08:50,900
the system.
137
00:08:50,900 --> 00:08:55,280
And if we apply the Laplace
transform to this equation and
138
00:08:55,280 --> 00:09:00,350
solve for the system function,
then the basic equation that
139
00:09:00,350 --> 00:09:03,870
we're left with expresses the
Laplace transform of the
140
00:09:03,870 --> 00:09:09,080
angle, equal to the system
function, the open-loop system
141
00:09:09,080 --> 00:09:13,220
function, times the Laplace
transform of
142
00:09:13,220 --> 00:09:14,750
the combined inputs.
143
00:09:14,750 --> 00:09:18,190
And I remind you again that our
assumption is that this is
144
00:09:18,190 --> 00:09:21,080
an input that we have
no control over.
145
00:09:21,080 --> 00:09:24,180
This is the input that
we can control.
146
00:09:24,180 --> 00:09:26,250
And the two together,
of course, form
147
00:09:26,250 --> 00:09:27,990
the combined input.
148
00:09:27,990 --> 00:09:31,650
Now, we can, of course, solve
for the poles and 0's.
149
00:09:31,650 --> 00:09:33,190
There are no 0's.
150
00:09:33,190 --> 00:09:35,310
And there are two poles, since
this is a second-order
151
00:09:35,310 --> 00:09:36,760
denominator.
152
00:09:36,760 --> 00:09:40,800
And looking at the poles in the
s-plane then, we see that
153
00:09:40,800 --> 00:09:44,750
we have a pair of poles, one at
minus the square root of g
154
00:09:44,750 --> 00:09:49,860
over L, and one at plus the
square root of g over L.
155
00:09:49,860 --> 00:09:54,390
And the important observation,
then, is that while this pole
156
00:09:54,390 --> 00:09:59,100
represents a stable pole, the
right half-plane pole
157
00:09:59,100 --> 00:10:01,470
represents an unstable pole.
158
00:10:01,470 --> 00:10:06,620
And so this system, in fact,
is an unstable system.
159
00:10:06,620 --> 00:10:11,170
It's unstable because what we
have is a pole in the right
160
00:10:11,170 --> 00:10:15,630
half-plane for the
open-loop system.
161
00:10:15,630 --> 00:10:19,750
Now, before we see how to
stabilize the system, let's,
162
00:10:19,750 --> 00:10:25,650
in fact, look at the mechanical
setup, which we
163
00:10:25,650 --> 00:10:30,540
won't turn on for now, and just
see essentially what this
164
00:10:30,540 --> 00:10:34,830
instability means and what the
physical orientation of the
165
00:10:34,830 --> 00:10:36,670
equipment is.
166
00:10:36,670 --> 00:10:40,090
So what we have, which you just
saw a caricature of in
167
00:10:40,090 --> 00:10:45,310
the view graph, is an
inverted pendulum.
168
00:10:45,310 --> 00:10:48,270
This represents the pendulum.
169
00:10:48,270 --> 00:10:52,550
And it's mounted with the pivot
point at the bottom.
170
00:10:52,550 --> 00:10:54,340
It's mounted on a cart.
171
00:10:54,340 --> 00:10:57,020
And here we have the cart.
172
00:10:57,020 --> 00:11:00,830
And this is a pivot point.
173
00:11:00,830 --> 00:11:04,100
And, as you can see, the
cart can move back
174
00:11:04,100 --> 00:11:06,110
and forth on a track.
175
00:11:06,110 --> 00:11:10,860
And the external acceleration,
which is applied to the cart,
176
00:11:10,860 --> 00:11:14,120
is applied through this cable.
177
00:11:14,120 --> 00:11:18,480
And that cable is controlled
by a motor.
178
00:11:18,480 --> 00:11:23,300
And we have the motor at
this end of the track.
179
00:11:23,300 --> 00:11:27,650
And so as the motor turns,
then the cart will
180
00:11:27,650 --> 00:11:29,760
move back and forth.
181
00:11:29,760 --> 00:11:30,250
OK.
182
00:11:30,250 --> 00:11:34,350
Now, since the system
is not turned on
183
00:11:34,350 --> 00:11:36,080
and there's no feedback--
184
00:11:36,080 --> 00:11:40,480
in fact, the system, as we just
saw in the transparency,
185
00:11:40,480 --> 00:11:42,710
is an unstable system.
186
00:11:42,710 --> 00:11:47,050
And what that instability means
is that, for example, if
187
00:11:47,050 --> 00:11:53,190
I set the angle to 0 and then
let it go, then as soon as
188
00:11:53,190 --> 00:11:55,310
there's a slight external
disturbance, it
189
00:11:55,310 --> 00:11:57,090
will start to fall.
190
00:11:57,090 --> 00:12:02,370
Now, you can imagine that not
only is the system unstable as
191
00:12:02,370 --> 00:12:06,530
it is, but certainly can't
accommodate changes in the
192
00:12:06,530 --> 00:12:07,690
system dynamics.
193
00:12:07,690 --> 00:12:09,660
For example, if we change,
let's say, the
194
00:12:09,660 --> 00:12:10,990
weight of the pendulum.
195
00:12:10,990 --> 00:12:15,960
So if we thought, for example,
of changing the weight by,
196
00:12:15,960 --> 00:12:19,350
let's say, putting something
like a glass with something in
197
00:12:19,350 --> 00:12:23,190
it on top, and I think about
trying to balance it by
198
00:12:23,190 --> 00:12:25,760
itself, obviously that's hard.
199
00:12:25,760 --> 00:12:29,030
In fact, I would say, without
turning the system on,
200
00:12:29,030 --> 00:12:31,710
guaranteed to be impossible.
201
00:12:31,710 --> 00:12:36,430
So the basic system, as we've
just analyzed it, is
202
00:12:36,430 --> 00:12:38,630
inherently an unstable system.
203
00:12:38,630 --> 00:12:42,860
The instability reflecting in
the fact that the pendulum,
204
00:12:42,860 --> 00:12:46,090
following its own natural
forces, will tend to fall.
205
00:12:46,090 --> 00:12:50,160
And now what we want to look at
is how, through the use of
206
00:12:50,160 --> 00:12:55,820
feedback, we can, in fact,
stabilize the system.
207
00:12:55,820 --> 00:13:05,160
So let's first, once again, look
at the open-loop system.
208
00:13:05,160 --> 00:13:09,410
And the open-loop system
function that we saw was a
209
00:13:09,410 --> 00:13:14,580
system function of this form, 1
over Ls squared minus g, and
210
00:13:14,580 --> 00:13:18,930
that represents the system
function associated with the
211
00:13:18,930 --> 00:13:22,390
two inputs, one being the
external disturbances, the
212
00:13:22,390 --> 00:13:25,290
other being the externally
applied acceleration.
213
00:13:25,290 --> 00:13:29,470
And the system function here
214
00:13:29,470 --> 00:13:31,200
represents the system dynamics.
215
00:13:31,200 --> 00:13:35,670
And the resulting output
is the angle.
216
00:13:35,670 --> 00:13:40,230
Now, the feedback, the basic
strategy behind the feedback,
217
00:13:40,230 --> 00:13:45,110
is to in some way use the
measured angle, make a
218
00:13:45,110 --> 00:13:49,000
measurement of the angle, and
use that through appropriate
219
00:13:49,000 --> 00:13:52,160
feedback dynamics to stabilize
the system.
220
00:13:52,160 --> 00:13:56,920
And so with feedback applied
around the system then, there
221
00:13:56,920 --> 00:14:00,770
would be some measurement of
the angle through some
222
00:14:00,770 --> 00:14:03,550
appropriately chosen
feedback dynamics.
223
00:14:03,550 --> 00:14:09,490
And that then would determine
for us the applied external
224
00:14:09,490 --> 00:14:13,420
acceleration corresponding to
what the motor does by pulling
225
00:14:13,420 --> 00:14:15,210
the cable back and forth.
226
00:14:15,210 --> 00:14:19,070
And we would like to choose
G of s so that the overall
227
00:14:19,070 --> 00:14:21,270
system is stable.
228
00:14:21,270 --> 00:14:24,980
Now, as you recall from the
lecture last time, with the
229
00:14:24,980 --> 00:14:28,890
feedback around the system and
expressed here in terms of
230
00:14:28,890 --> 00:14:34,480
negative feedback, the overall
transfer function then is the
231
00:14:34,480 --> 00:14:38,030
transfer function associated
with the basic feedback loop,
232
00:14:38,030 --> 00:14:43,420
where H of s is the open-loop
system and G of s corresponds
233
00:14:43,420 --> 00:14:46,160
to the feedback dynamics.
234
00:14:46,160 --> 00:14:49,070
Now, that's the basic
strategy.
235
00:14:49,070 --> 00:14:52,490
We haven't decided yet how to
choose the feedback dynamics.
236
00:14:52,490 --> 00:14:54,710
And that becomes
the next step.
237
00:14:54,710 --> 00:14:57,500
And we want to choose the
dynamics in such a way that
238
00:14:57,500 --> 00:15:01,840
the system ends up
being stabilized.
239
00:15:01,840 --> 00:15:06,290
Well, I think the thing to do is
just simply begin with what
240
00:15:06,290 --> 00:15:10,060
would seem to be the most
obvious, which is a simple
241
00:15:10,060 --> 00:15:11,750
measurement of the angle.
242
00:15:11,750 --> 00:15:16,910
Let's take the angular
measurements, feed that back,
243
00:15:16,910 --> 00:15:19,770
let's say through a
potentiometer and perhaps an
244
00:15:19,770 --> 00:15:22,850
amplifier, so that there's some
gain, and see if we can
245
00:15:22,850 --> 00:15:27,410
stabilize the system simply
through feedback which is
246
00:15:27,410 --> 00:15:30,160
proportional to a measurement
of the angle, what.
247
00:15:30,160 --> 00:15:36,710
Is typically referred to as
"proportional feedback." Well,
248
00:15:36,710 --> 00:15:40,200
let's analyze the results
of doing that.
249
00:15:40,200 --> 00:15:46,380
Once again we have the basic
feedback equation, where the
250
00:15:46,380 --> 00:15:50,730
open-loop transfer function
is what we had developed
251
00:15:50,730 --> 00:15:54,670
previously, given by the
second-order expression.
252
00:15:54,670 --> 00:15:59,570
Now, using proportional
feedback, we choose the
253
00:15:59,570 --> 00:16:05,145
acceleration a of t to be
directly proportional through
254
00:16:05,145 --> 00:16:09,260
a gain constant or attenuation
constant K1, directly
255
00:16:09,260 --> 00:16:12,690
proportional to the measured
angle theta of t.
256
00:16:12,690 --> 00:16:16,970
And consequently the system
function in the feedback path
257
00:16:16,970 --> 00:16:20,400
is just simply a gain
or attenuate K1.
258
00:16:20,400 --> 00:16:23,270
So this, then, is the
system function for
259
00:16:23,270 --> 00:16:25,270
the feedback path.
260
00:16:25,270 --> 00:16:31,720
Well, substituting this into
the closed-loop expression,
261
00:16:31,720 --> 00:16:36,970
then the overall expression for
the Laplace transform of
262
00:16:36,970 --> 00:16:42,750
the output angle is, as I
indicate here, namely that
263
00:16:42,750 --> 00:16:48,170
theta of s is proportional
through this system function.
264
00:16:48,170 --> 00:16:52,570
So the Laplace transform of the
input X of t-- and let me
265
00:16:52,570 --> 00:16:56,930
remind you that X of t, the
external disturbances, now
266
00:16:56,930 --> 00:17:00,040
represent the only input,
since the other input
267
00:17:00,040 --> 00:17:03,300
corresponding to the applied
acceleration to the cart is
268
00:17:03,300 --> 00:17:06,930
now controlled only through
the feedback loop.
269
00:17:06,930 --> 00:17:09,950
So we have this system function,
then, for the
270
00:17:09,950 --> 00:17:11,579
closed-loop system.
271
00:17:11,579 --> 00:17:15,990
We recognize this once again
as a second-order system.
272
00:17:15,990 --> 00:17:20,960
And the poles of this
second-order system are then
273
00:17:20,960 --> 00:17:25,339
given by plus and minus the
square root of g minus K1,
274
00:17:25,339 --> 00:17:31,550
divided by L. So K1, the
feedback constant, clearly
275
00:17:31,550 --> 00:17:34,320
influences the position
of the poles.
276
00:17:34,320 --> 00:17:37,960
And let's look, in fact,
at where the
277
00:17:37,960 --> 00:17:42,200
poles are in the s-plane.
278
00:17:42,200 --> 00:17:48,090
Well, first of all, with K1
equal to 0, which, of course,
279
00:17:48,090 --> 00:17:52,290
is no feedback and corresponds
to the open-loop system, we
280
00:17:52,290 --> 00:17:56,480
have the poles where they
were previously.
281
00:17:56,480 --> 00:18:03,510
And if we now choose K1, let's
say less than 0, then what
282
00:18:03,510 --> 00:18:09,690
will happen as K1 becomes more
and more negative, so that
283
00:18:09,690 --> 00:18:12,965
this term is more and more
positive, is that the left
284
00:18:12,965 --> 00:18:16,425
half-plane pole will move
further into the left
285
00:18:16,425 --> 00:18:20,150
half-plane, the right half-plane
pole will move
286
00:18:20,150 --> 00:18:22,560
further into the right
half-plane.
287
00:18:22,560 --> 00:18:28,960
So clearly with K1 negative,
this pole, which represents an
288
00:18:28,960 --> 00:18:34,010
instability, becomes
even more unstable.
289
00:18:34,010 --> 00:18:38,450
Well, instead of K1 negative,
let's try K1 positive and see
290
00:18:38,450 --> 00:18:39,740
what happens.
291
00:18:39,740 --> 00:18:47,890
And with K1 positive, what
happens in this case is that
292
00:18:47,890 --> 00:18:53,900
the left half-plane pole moves
closer to the origin, the
293
00:18:53,900 --> 00:18:57,420
right half-plane pole moves
closer to the origin.
294
00:18:57,420 --> 00:19:00,360
What one would hope is that they
both end up in the left
295
00:19:00,360 --> 00:19:01,680
half-plane at some point.
296
00:19:01,680 --> 00:19:03,450
But, in fact, they don't.
297
00:19:03,450 --> 00:19:06,370
What happens is that eventually
they both reach the
298
00:19:06,370 --> 00:19:11,010
origin, split at that point,
and travel along the
299
00:19:11,010 --> 00:19:13,360
j-omega-axis.
300
00:19:13,360 --> 00:19:19,020
Now, this movement of the poles,
as we vary K1 either
301
00:19:19,020 --> 00:19:22,350
positive or negative, what we
see is that with the open-loop
302
00:19:22,350 --> 00:19:26,760
system, as we introduce
feedback, those basic poles
303
00:19:26,760 --> 00:19:31,080
move in the s-plane, either this
way, as K1 becomes more
304
00:19:31,080 --> 00:19:35,010
negative, or for this particular
case, if K1 is
305
00:19:35,010 --> 00:19:37,300
positive, they move in together,
split, and move
306
00:19:37,300 --> 00:19:39,330
along the j-omega-axis.
307
00:19:39,330 --> 00:19:44,740
And that locus of the poles,
in fact, is referred to in
308
00:19:44,740 --> 00:19:48,540
feedback terminology as the
"root locus." And as is
309
00:19:48,540 --> 00:19:52,510
discussed in much more detail in
the text, there are lots of
310
00:19:52,510 --> 00:19:57,580
ways of determining the root
locus for feedback systems
311
00:19:57,580 --> 00:20:00,470
without explicitly solving
for the roots.
312
00:20:00,470 --> 00:20:03,700
For this particular example, in
fact, we can determine the
313
00:20:03,700 --> 00:20:07,420
root locus in the most
straightforward way simply by
314
00:20:07,420 --> 00:20:10,060
solving for the roots
of the denominator
315
00:20:10,060 --> 00:20:11,810
of the system function.
316
00:20:11,810 --> 00:20:17,380
Now, notice that in this case,
these poles, with K1 positive,
317
00:20:17,380 --> 00:20:18,760
have moved together.
318
00:20:18,760 --> 00:20:21,570
They move along the
j-omega-axis.
319
00:20:21,570 --> 00:20:24,850
And before they come
together, the
320
00:20:24,850 --> 00:20:27,460
system is clearly unstable.
321
00:20:27,460 --> 00:20:31,510
Even when they come together
and split, the system is
322
00:20:31,510 --> 00:20:35,190
marginally stable, because,
in fact, the system
323
00:20:35,190 --> 00:20:36,660
would tend to oscillate.
324
00:20:36,660 --> 00:20:39,420
And what that oscillation
means is that with the
325
00:20:39,420 --> 00:20:43,020
measurement of the angles,
essentially if the poles are
326
00:20:43,020 --> 00:20:48,230
operating on the j-omega-axis,
what will happen is that
327
00:20:48,230 --> 00:20:50,460
things will oscillate
back and forth.
328
00:20:50,460 --> 00:20:54,210
Perhaps with the cart moving
back and forth trying to
329
00:20:54,210 --> 00:20:56,750
compensate, and the rod
sort of moving in
330
00:20:56,750 --> 00:20:58,870
the opposite direction.
331
00:20:58,870 --> 00:21:02,770
In any case, what's happened
is, with just proportional
332
00:21:02,770 --> 00:21:07,030
feedback, we apparently
are unable to
333
00:21:07,030 --> 00:21:10,250
stabilize the system.
334
00:21:10,250 --> 00:21:13,140
Well, you could think that the
reason, perhaps, is that we're
335
00:21:13,140 --> 00:21:14,490
not responding fast enough.
336
00:21:14,490 --> 00:21:19,310
For example, if the angle starts
to change, perhaps, in
337
00:21:19,310 --> 00:21:23,540
fact, we should make the
feedback proportional to the
338
00:21:23,540 --> 00:21:25,780
rate of change of angles,
rather than
339
00:21:25,780 --> 00:21:27,280
to the angle itself.
340
00:21:27,280 --> 00:21:30,710
And so we could examine the
possibility of using what's
341
00:21:30,710 --> 00:21:34,700
referred to as "derivative
feedback."
342
00:21:34,700 --> 00:21:42,770
In derivative feedback, what we
would do is to choose, for
343
00:21:42,770 --> 00:21:49,390
the feedback equation, or for
the feedback system function,
344
00:21:49,390 --> 00:21:53,020
something reflecting a
measurement of the derivative
345
00:21:53,020 --> 00:21:54,410
of the angle.
346
00:21:54,410 --> 00:21:58,910
And so here again, once again,
we have the open-loop system
347
00:21:58,910 --> 00:22:01,490
function with derivative
feedback.
348
00:22:01,490 --> 00:22:03,820
We'll attempt to--
349
00:22:03,820 --> 00:22:06,040
or we will use this feedback.
350
00:22:06,040 --> 00:22:09,040
And acceleration, which instead
of being proportional
351
00:22:09,040 --> 00:22:12,540
to the angle, as it was in the
previous case, an acceleration
352
00:22:12,540 --> 00:22:16,840
which is proportional to the
derivative of the angle.
353
00:22:16,840 --> 00:22:20,470
And so the basic feedback
dynamics, or feedback system
354
00:22:20,470 --> 00:22:25,060
function, is then G of s
is equal to K2 times s.
355
00:22:25,060 --> 00:22:28,110
The multiplication by s
reflecting the fact that in
356
00:22:28,110 --> 00:22:33,130
the time domain it's measuring
the derivative of the angle.
357
00:22:33,130 --> 00:22:38,670
The associated system function
is then indicated here.
358
00:22:38,670 --> 00:22:42,960
And if we solve again this
second-order equation for its
359
00:22:42,960 --> 00:22:47,730
roots, that tells us that the
location of the poles are
360
00:22:47,730 --> 00:22:51,810
given by this equation.
361
00:22:51,810 --> 00:22:57,100
And so now what we would want to
look at is how these poles
362
00:22:57,100 --> 00:23:03,950
move as we vary the derivative
feedback constant K2.
363
00:23:03,950 --> 00:23:08,810
So let's look at the root
locus for that case.
364
00:23:08,810 --> 00:23:17,860
And first, once again, we have
the basic open-loop poles.
365
00:23:17,860 --> 00:23:20,915
And the open-loop poles consist
of a pole on the left
366
00:23:20,915 --> 00:23:23,880
half-plane and a pole on
the right half-plane,
367
00:23:23,880 --> 00:23:28,400
corresponding in this equation
to K2 equal to 0.
368
00:23:28,400 --> 00:23:32,700
That is, no feedback
in the system.
369
00:23:32,700 --> 00:23:37,840
If K2 is negative, then what you
can see is that this real
370
00:23:37,840 --> 00:23:41,700
part will become
more negative.
371
00:23:41,700 --> 00:23:46,220
Since K2 is squared here, this
is still a positive quantity.
372
00:23:46,220 --> 00:23:51,280
And, in fact, then with K2 less
than 0, the root locus
373
00:23:51,280 --> 00:23:55,050
that we get is indicated
by this.
374
00:23:55,050 --> 00:23:59,990
Now, notice, then, that this
right half-plane pole is
375
00:23:59,990 --> 00:24:02,020
becoming more unstable.
376
00:24:02,020 --> 00:24:04,460
And the left half-plane
pole likewise is
377
00:24:04,460 --> 00:24:06,250
becoming more unstable.
378
00:24:06,250 --> 00:24:09,800
And this point, by the way,
corresponds to where this pole
379
00:24:09,800 --> 00:24:13,930
ends up, or where the root
locus ends up, when K2
380
00:24:13,930 --> 00:24:15,680
eventually becomes infinite.
381
00:24:15,680 --> 00:24:20,450
So clearly K2 negative is not
going to stabilize the system.
382
00:24:20,450 --> 00:24:24,700
Let's try K2 positive.
383
00:24:24,700 --> 00:24:29,960
And the root locus dictated by
this equation, then, is what I
384
00:24:29,960 --> 00:24:31,810
indicate here.
385
00:24:31,810 --> 00:24:35,130
The left half-plane pole moves
further into the left
386
00:24:35,130 --> 00:24:35,760
half-plane.
387
00:24:35,760 --> 00:24:36,340
That's good.
388
00:24:36,340 --> 00:24:38,430
That's getting more stable.
389
00:24:38,430 --> 00:24:42,600
The right half-plane pole is
moving closer to the left
390
00:24:42,600 --> 00:24:46,630
half-plane, but unfortunately
never gets there.
391
00:24:46,630 --> 00:24:50,840
And, in fact, it's when K2
eventually becomes infinite
392
00:24:50,840 --> 00:24:54,600
that this pole just gets to
the point where the system
393
00:24:54,600 --> 00:24:57,870
becomes marginally stable.
394
00:24:57,870 --> 00:25:02,010
So what we found is that with
proportional feedback, we
395
00:25:02,010 --> 00:25:05,640
can't stabilize the system, with
the derivative feedback,
396
00:25:05,640 --> 00:25:09,930
we can't stabilize the system,
by themselves.
397
00:25:09,930 --> 00:25:13,890
And a logical next choice is to
see if we can stabilize the
398
00:25:13,890 --> 00:25:19,790
system by both measuring the
angle and at the same time
399
00:25:19,790 --> 00:25:23,440
being careful to be responsive
to how fast that angle is
400
00:25:23,440 --> 00:25:27,340
changing, so that if it's
changing too fast, we can move
401
00:25:27,340 --> 00:25:30,630
the cart or our hand
under the inverted
402
00:25:30,630 --> 00:25:32,950
pendulum more quickly.
403
00:25:32,950 --> 00:25:37,370
Well, now then what we want
to examine is the use of
404
00:25:37,370 --> 00:25:40,290
proportional plus derivative
feedback.
405
00:25:40,290 --> 00:25:49,610
And in that case, we then have a
choice for the acceleration,
406
00:25:49,610 --> 00:25:54,330
which is proportional with one
constant to the angle and
407
00:25:54,330 --> 00:25:57,360
proportional with another
constant to the
408
00:25:57,360 --> 00:25:59,350
derivative of the angle.
409
00:25:59,350 --> 00:26:04,480
And so the basic system
function, then, with
410
00:26:04,480 --> 00:26:07,850
proportional plus derivative
feedback, is a system function
411
00:26:07,850 --> 00:26:12,330
which is K1 plus K2 times s.
412
00:26:12,330 --> 00:26:17,280
We then have an overall
closed-loop system function,
413
00:26:17,280 --> 00:26:23,870
theta of s, which is given
by this equation.
414
00:26:23,870 --> 00:26:28,900
And so the roots of this
equation then represent the
415
00:26:28,900 --> 00:26:32,470
poles of the closed-loop
system.
416
00:26:32,470 --> 00:26:37,980
And those poles involve
two parameters.
417
00:26:37,980 --> 00:26:42,120
They involve the parameter K2,
which is proportional to the
418
00:26:42,120 --> 00:26:45,740
derivative of the angle, and
the constant K1, which is
419
00:26:45,740 --> 00:26:47,700
proportional to the angle.
420
00:26:47,700 --> 00:26:55,160
And we'll, first of all, examine
this just with K2
421
00:26:55,160 --> 00:26:59,560
positive, because what we can
see is that as we vary K1, if
422
00:26:59,560 --> 00:27:05,120
K2 were negative, that would,
more or less immediately, put
423
00:27:05,120 --> 00:27:07,270
poles into the right
half-plane.
424
00:27:07,270 --> 00:27:11,280
And the more negative K2
got, the larger this
425
00:27:11,280 --> 00:27:12,910
term is going to get.
426
00:27:12,910 --> 00:27:15,790
So, in fact, as it will turn
out, we can stabilize the
427
00:27:15,790 --> 00:27:20,350
system, provided that
we choose K2 to be
428
00:27:20,350 --> 00:27:23,160
greater than 0.
429
00:27:23,160 --> 00:27:23,410
All right.
430
00:27:23,410 --> 00:27:30,960
Now, with K2 greater than 0,
what happens in the location
431
00:27:30,960 --> 00:27:37,640
of the poles is that if K2 is
greater than 0, and we choose
432
00:27:37,640 --> 00:27:39,330
K1 greater than 0--
433
00:27:39,330 --> 00:27:39,780
I'm sorry.
434
00:27:39,780 --> 00:27:45,870
We choose K1 equal to 0, then in
effect the influence of K2
435
00:27:45,870 --> 00:27:51,220
is to shift the poles of the
open-loop system slightly.
436
00:27:51,220 --> 00:27:58,450
And so with K2 greater than 0
and K1 equal to 0, we have a
437
00:27:58,450 --> 00:28:06,250
set of poles, which are
indicated here, and so this is
438
00:28:06,250 --> 00:28:08,800
just a shift, slight shift, to
the left, depending on the
439
00:28:08,800 --> 00:28:13,900
value of K2, a shift to the left
of the open-loop poles.
440
00:28:13,900 --> 00:28:17,190
Now, as we vary K1, and in
particular we're going to
441
00:28:17,190 --> 00:28:24,700
choose K1 greater than 0, what
happens is that the poles will
442
00:28:24,700 --> 00:28:31,000
begin to move together, as they
did previously when we
443
00:28:31,000 --> 00:28:33,230
looked at the variation of K1.
444
00:28:33,230 --> 00:28:36,500
The poles will move together,
reach a point where we have a
445
00:28:36,500 --> 00:28:41,820
second-order pole, and then
those poles will split and
446
00:28:41,820 --> 00:28:43,900
move parallel to the
j-omega-axis.
447
00:28:43,900 --> 00:28:48,100
So what's indicated here
is the root locus.
448
00:28:48,100 --> 00:28:52,990
And what this represents, then,
as long as we make K1
449
00:28:52,990 --> 00:28:56,075
large enough so that this poles
moves into the left
450
00:28:56,075 --> 00:28:59,570
half-plane, is that
it represents
451
00:28:59,570 --> 00:29:02,050
now a stable system.
452
00:29:02,050 --> 00:29:02,330
OK.
453
00:29:02,330 --> 00:29:05,840
So what we've seen is that
proportional feedback by
454
00:29:05,840 --> 00:29:10,330
itself or derivative feedback by
itself won't stabilize the
455
00:29:10,330 --> 00:29:14,820
system, whereas with the right
choice of feedback constants,
456
00:29:14,820 --> 00:29:16,900
proportional plus derivative
feedback will.
457
00:29:16,900 --> 00:29:20,620
And we saw that basically by
examining the root locus in
458
00:29:20,620 --> 00:29:22,080
the s-plane.
459
00:29:22,080 --> 00:29:22,350
All right.
460
00:29:22,350 --> 00:29:26,200
Well, let's actually watch
the system in action.
461
00:29:26,200 --> 00:29:29,820
And I described it to
you previously.
462
00:29:29,820 --> 00:29:33,730
Basically, an inverted
pendulum on a cart.
463
00:29:33,730 --> 00:29:36,630
And so I still have it off.
464
00:29:36,630 --> 00:29:38,870
And, of course, we have
the pendulum.
465
00:29:38,870 --> 00:29:43,310
And as I indicated, it's
pivoted at the base.
466
00:29:43,310 --> 00:29:50,260
And the angle is measured by a
potentiometer that we have
467
00:29:50,260 --> 00:29:52,240
attached to the pivot point.
468
00:29:52,240 --> 00:29:57,040
And the measurement of the angle
is fed back through this
469
00:29:57,040 --> 00:30:02,370
wire to a motor that we have at
the other end of the table.
470
00:30:02,370 --> 00:30:06,350
And then that motor basically
is used to provide the
471
00:30:06,350 --> 00:30:09,660
acceleration to drive
the cart.
472
00:30:09,660 --> 00:30:09,950
OK.
473
00:30:09,950 --> 00:30:12,330
Well, let's turn it on.
474
00:30:12,330 --> 00:30:15,120
And when we turn it on,
it'll take just
475
00:30:15,120 --> 00:30:16,370
an instant to stabilize.
476
00:30:16,370 --> 00:30:19,030
477
00:30:19,030 --> 00:30:24,200
And fortunately we have the
constants set right, and there
478
00:30:24,200 --> 00:30:28,710
we have now the stabilization
of an unstable system.
479
00:30:28,710 --> 00:30:31,460
Remember that with the feedback
off, the system is
480
00:30:31,460 --> 00:30:34,610
unstable because the pendulum
will fall, whereas now it's
481
00:30:34,610 --> 00:30:35,760
stabilized.
482
00:30:35,760 --> 00:30:41,680
Now, also, as you can see, not
only have we stabilized it,
483
00:30:41,680 --> 00:30:45,340
but we're able to compensate
through the feedback to
484
00:30:45,340 --> 00:30:47,540
changes in the external
disturbances.
485
00:30:47,540 --> 00:30:51,490
For example, by tapping it,
because of the feedback and
486
00:30:51,490 --> 00:30:54,200
the measurement of the angle,
it will more or less
487
00:30:54,200 --> 00:30:56,880
automatically stabilize.
488
00:30:56,880 --> 00:31:00,600
Now, in addition to being stable
in the presence of
489
00:31:00,600 --> 00:31:04,940
external disturbances, it also
remains stable and remains
490
00:31:04,940 --> 00:31:09,210
balanced even if we were to
change the system dynamics.
491
00:31:09,210 --> 00:31:13,950
And let me just illustrate that
with the glass that we've
492
00:31:13,950 --> 00:31:15,510
talked about before.
493
00:31:15,510 --> 00:31:19,340
Let's first not be too bold, and
we'll take the liquid out
494
00:31:19,340 --> 00:31:20,420
of the glass.
495
00:31:20,420 --> 00:31:24,440
And presumably if it can adjust
to changes in the
496
00:31:24,440 --> 00:31:28,850
system dynamics, then if I put
the glass on, in fact, it will
497
00:31:28,850 --> 00:31:29,660
remain balanced.
498
00:31:29,660 --> 00:31:31,220
And indeed it does.
499
00:31:31,220 --> 00:31:33,890
And let me point out, by the
way, that I don't have to be
500
00:31:33,890 --> 00:31:37,480
very careful about exactly where
I position the glass.
501
00:31:37,480 --> 00:31:41,770
And furthermore, I can change
the overall system even
502
00:31:41,770 --> 00:31:47,480
further by, let's say for
example, pouring a liquid in.
503
00:31:47,480 --> 00:31:51,540
And now let me also comment that
I've changed the physics
504
00:31:51,540 --> 00:31:52,200
of it a little bit.
505
00:31:52,200 --> 00:31:55,920
Because the liquid can slosh
around a little bit, it
506
00:31:55,920 --> 00:31:58,440
becomes a little more
complicated a system.
507
00:31:58,440 --> 00:32:03,650
But as you can see, it still
remains balanced.
508
00:32:03,650 --> 00:32:08,270
Now, if we really don't want to
be too conservative at all,
509
00:32:08,270 --> 00:32:11,022
we could wonder whether, with
the feedback constants we
510
00:32:11,022 --> 00:32:14,290
have, we could, in
fact, balance the
511
00:32:14,290 --> 00:32:15,450
pitcher on the top.
512
00:32:15,450 --> 00:32:21,000
And, well, I guess we may
as well give that a try.
513
00:32:21,000 --> 00:32:27,270
And so now we're changing the
mass at the top of the
514
00:32:27,270 --> 00:32:31,200
pendulum by a considerable
amount.
515
00:32:31,200 --> 00:32:34,890
And, again, the system basically
can respond to it.
516
00:32:34,890 --> 00:32:38,090
Now, this is a fairly
complicated system.
517
00:32:38,090 --> 00:32:40,420
The liquid is sloshing around.
518
00:32:40,420 --> 00:32:44,210
We, in fact, as you can see,
have an instability right now,
519
00:32:44,210 --> 00:32:45,770
although it's controlled.
520
00:32:45,770 --> 00:32:49,060
And that's because the physics
of the dynamics has changed.
521
00:32:49,060 --> 00:32:53,130
And we can put a little bit more
mass into the system, and
522
00:32:53,130 --> 00:32:55,505
maybe or maybe not that will cut
down on the instability.
523
00:32:55,505 --> 00:33:02,060
524
00:33:02,060 --> 00:33:02,300
OK.
525
00:33:02,300 --> 00:33:06,050
Well, in fact, what happened
there is that we increased the
526
00:33:06,050 --> 00:33:09,150
mass at the top of the
pendulum slightly.
527
00:33:09,150 --> 00:33:12,020
And that provided just
enough damping to
528
00:33:12,020 --> 00:33:13,270
stabilize the system.
529
00:33:13,270 --> 00:33:16,480
530
00:33:16,480 --> 00:33:17,030
OK.
531
00:33:17,030 --> 00:33:22,870
Well, with this lecture and
this demonstration, this
532
00:33:22,870 --> 00:33:25,310
concludes this entire
set of lectures.
533
00:33:25,310 --> 00:33:28,270
It concludes it especially if
the pitcher happens to fall.
534
00:33:28,270 --> 00:33:34,040
But, seriously, this concludes
the set of lectures as we have
535
00:33:34,040 --> 00:33:36,180
put them together.
536
00:33:36,180 --> 00:33:41,880
And let me just comment that,
as a professor of mine once
537
00:33:41,880 --> 00:33:45,750
said, and which I've never
forgotten, the purpose of a
538
00:33:45,750 --> 00:33:48,470
set of lectures, or of a course,
or, for that matter
539
00:33:48,470 --> 00:33:53,260
anything that you study, is not
really to cover a subject,
540
00:33:53,260 --> 00:33:55,760
but to uncover the subject.
541
00:33:55,760 --> 00:34:01,980
And I hope that, at least to
some degree, we were able to
542
00:34:01,980 --> 00:34:05,020
uncover the topic of signals
and systems through this
543
00:34:05,020 --> 00:34:07,240
series of lectures.
544
00:34:07,240 --> 00:34:12,679
There are a lot of topics that
we got only a very brief
545
00:34:12,679 --> 00:34:14,010
glimpse into.
546
00:34:14,010 --> 00:34:18,290
And I hope that at least what
we've been able to do is get
547
00:34:18,290 --> 00:34:21,280
you interested enough in them so
that you'll pursue some of
548
00:34:21,280 --> 00:34:23,020
these on your own.
549
00:34:23,020 --> 00:34:27,699
And so I'd like to conclude by
thanking you, both for your
550
00:34:27,699 --> 00:34:29,159
patience and your interest.
551
00:34:29,159 --> 00:34:33,130
And I hope that you have enough
interest to pursue some
552
00:34:33,130 --> 00:34:34,469
of these topics further.
553
00:34:34,469 --> 00:34:35,200
Thank you.
554
00:34:35,200 --> 00:34:36,760
FILMING DIRECTOR: Okay.
555
00:34:36,760 --> 00:34:38,010
That's a wrap.
556
00:34:38,010 --> 00:34:41,920